Question

A single conservative force F(x) = b x + a acts on a 4.37 kg particle, where x is in meters, b = 6.09 N/m and a = 5.96 N. As the particle moves along the x axis from x1 = 0.652 m to x2 = 5.1 m, calculate the work done by this force. Answer in units of J.

Answer 1

Answer:104.41 J

Step-by-step explanation:

Given

Force acting
`F(x)=bx+a`

mass of particle
`m=4.37 kg`

`a=5.96 N`

`b=6.09 N/m`

Work done is given by

`W=\int_(a)^(b)F.dx`

`W=\int_(0.652)^(5.1)\left ( 6.09x+5.96\right )dx`

`W=\left [ 6.09* \left ( (x^2)/(2)\right )+5.96* x\right ]_(0.652)^(5.1)`

`W=77.906+26.51 J`

`W=104.41 J`